How do you find the rectangular equation for #theta=(5pi)/6#?

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Answer 1 · 2016-10-01T05:17:37

#y=-x/sqrt 3#, limited to the half-line in #Q_2#.

The conversion formula is #(r(cos theta, sim theta ) = (x, y, )#, giving

#x = r cos theta, y = r sin theta ade r =sqrt(x^2+y^2)>=0#.

Here #theta =5/6pi# represents the half-line from pole in the

direction in #Q_2#.

Sans #r = 0,

y/x= sin(5/6pi)/cos(5/6pi)#

#= sin(pi-pi/6)/cos(pi-pi/6)#

#=sin(pi/6)/(-cos(pi/6)#

#=-1/sqrt 3.

As a matter of fact, this half line is discontinuous at its end (0, 0).